Q.1. If x^{n} – 1 is divisible by x – k, then the least positive integral value of k is
(a) 1
(b) 2
(c) 3
(d) 4
Correct Answer is option (a)
Given,
P(n): xn – 1 is divisible by x – k
Let us substitute n = 1, 2, 3,..
⇒ P(1) : x – 1
⇒ P(2) : x^{2} – 1 = (x−1)(x+1)
⇒ P(3) : x^{3} – 1 = (x − 1)(x^{2 }+ x + 1)
⇒ P(4) : x^{4} − 1 = (x^{2} − 1)(x^{2} + 1) = (x − 1)(x + 1)(x^{2} + 1)
Therefore, the least positive integral value of k is 1.
Q.2. The coefficient of the middle term in the expansion of (2+3x)^{4 }is:
(a) 5!
(b) 6
(c) 216
(d) 8!
Correct Answer is option (c)
If the exponent of the expression is n, then the total number of terms is n+1.
Hence, the total number of terms is 4+1 = 5.
Hence, the middle term is the 3rd term.
We know that general term of (x+a)^{n} is T_{r+1} =^{n}C_{r }x^{nr }a^{r}
Here, n=4, r=2
Therefore, T_{3} = ^{4}C_{2}.(2)^{2}.(3x)^{2}
T_{3} = (6).(4).(9x^{2})
T_{3} = 216x^{2}.
Therefore, the coefficient of the middle term is 216.
Q.3. For any natural number n, 2^{2n} – 1 is divisible b
(a) 2
(b) 3
(c) 4
(d) 5
Correct Answer is option (b)
Let P(n) = 2^{2n} – 1
Substituting n = 1, 2, 3,….
P(1) = 2^{2(1)} – 1 = 4 – 1 = 3
This is divisible by 3.
P(2) = 2^{2(2)} – 1 = 16 – 1 = 15
This is divisible by 3.
P(3) = 2^{2(3)} – 1 = 256 – 1 = 255
This is also divisible by 3.
Assume that P(n) is true for some natural number k, i.e., P(k): 2^{2k} – 1 is divisible by 3, i.e., 2^{2k} – 1
= 3q, where q ∈ N
Now,
P(k + 1) : 2^{2(k+1)} – 1
= 2^{2k + 2} – 1
= 2^{2k} . 2^{2} – 1
= 2^{2k} . 4 – 1
= 3.2^{2k} + (2^{2k} – 1)
= 3.2^{2k} + 3q
= 3 (2^{2k} + q) = 3m, where m ∈ N
Thus P(k + 1) is true, whenever P(k) is true.
Therefore, for any natural number n, 2^{2n} – 1 is divisible by 3.
Q.4. The value of (126)^{1/3} up to three decimal places is
(a) 5.011
(b) 5.012
(c) 5.013
(d) 5.014
Correct Answer is option (c)
(126)^{⅓ }can also be written as the cube root of 126.
Hence, (126)^{⅓} is approximately equal to 5.013.
Hence, option (c) 5.013 is the correct answer.
Q.5. A student was asked to prove a statement P(n) by induction. He proved that P(k + 1) is true whenever P(k) is true for all k > 5 ∈ N and also that P(5) is true. Based on this, he could conclude that P(n) is true
(a) for all n ∈ N
(b) for all n > 5
(c) for all n ≥ 5
(d) for all n < 5
Correct Answer is option (c)
The student could be able to conclude that P(n) is true for all n ≥ 5 since P(5) is true for all k > 5 ∈ N as well as true for P(5) and P(k + 1) is true, whenever P(k) is true.
Q.6. If n is even in the expansion of (a+b)^{n}, the middle term is:
(a) n^{th} term
(b) (n/2)^{th} term
(c) [(n/2)1]^{th} term
(d) [(n/2)+1]^{th} term
Correct Answer is option (d)
In general, if “n” is the even in the expansion of (a+b)^{n}, then the number of terms will be odd. (i.e) n+1.
Hence, the middle term of the expansion (a+b)^{n} is [(n/2)+1]^{th} term.
Q.7. If P (n): “49^{n} + 16^{n} + k is divisible by 64 for n ∈ N” is true, then the least negative integral value of k is
(a) 1
(b) 2
(c) 1
(d) 3
Correct Answer is option (c)
Given that P(n) : 49^{n} + 16^{n} + k is divisible by 64 for n ∈ N
For n = 1,
P(1) : 49 + 16 + k = 65 + k is divisible by 64.
Thus k, should be 1 since, 65 – 1 = 64 is divisible by 64.
Q.8. The largest coefficient in the expansion of (1 + x)^{10} is:
(a) 10! / (5!)^{2}
(b) 10! / 5!
(c) 10! / (5!×4!)^{2}
(d) 10! / (5!×4!)
Correct Answer is option (a)
Given: (1+x)^{10}
The greatest coefficient will always occur in the middle term.
Hence, the total number of terms in an expansion is 11. (i.e. 10+1 = 11)
Therefore, middle term = [(10/2) + 1] = 5+1 = 6th term.
We know that general term of (x+a)^{n} is T_{r+1} =^{n}C_{r }x^{nr }a^{r}
Here, n=10, r=5
So, T6 = ^{10}C_{5}.x^{5}
Therefore, the coefficient of the greatest term = ^{10}C_{5} = 10!/(5!)^{2}.
So, option (a) 10!/(5!)^{2} is the correct answer.
Q.9. For all n ∈ N, 3.5^{2n+1} + 2^{3n+1} is divisible by
(a) 19
(b) 17
(c) 23
(d) 25
Correct Answer is option (b)
Let P(n) be the statement that 3.5^{2n + 1} + 2^{3n + 1} is divisible by 17
If n = 1, then given expression = 3 * 5^{3} + 2^{4} + 375 + 16 = 391 = 17 * 23, divisible by 17.
P(1) is true
Assume that P(k) is true.
3.5^{2k + 1} + 2^{3k + 1} is divisible by 17.
3.5^{2k} = 1 + 2^{3k + 1} = 17m where m ∈ N
3.5^{2(k + 1) + 1} + 23^{(k + 1) + 1}
= 3.5^{2k + 1} * 5^{2} + 2^{3k + 1} * 2^{3}
= 25^{(17m – 23k+1)} + 8.2^{3k + 1}
= 425m – 25.2^{3k + 1} + 8.2^{3k + 1}
= 425m – 17.2^{3k + 1}
= 17(25m – 2^{3k + 1}), divisible by 17
P(k + 1) is true by Principle of Mathematical Induction
P(n) is true for all n ∈ N. 3.5^{2n + 1} + 2^{3n + 1} is divisible by 17 for all n ∈ N
Q.10. The coefficient of x^{3}y^{4} in (2x+3y^{2})^{5} is
(a) 360
(b) 720
(c) 240
(d) 1080
Correct Answer is option (b)
Given: (2x+3y^{2})^{5}
Therefore, the general form for the expression (2x+3y^{2})^{5} is T_{r+1} = ^{5}C_{r}. (2x)^{r}.(3y^{2})^{5r}
Hence, T_{3+1} = ^{5}C_{3} (2x)^{3}.(3y^{2})^{53}
T_{4} = ^{5}C_{3} (2x)^{3}.(3y^{2})^{2}
T_{4} = ^{5}C_{3}.8x^{3}.9y^{4}
On simplification, we get
T_{4} = 720x^{3}y^{4}
Therefore, the coefficient of x^{3}y^{4} in (2x+3y^{2})^{5} is 720.
Q.11. n(n + 1) (n + 5) is a multiple of
(a) 3
(b) 8
(c) 5
(d) 7
Correct Answer is option (a)
Let P(n) = n(n + 1)(n + 5)
Substituting n = 1, 2, 3,….
P(1) = 1(1 + 1)(1 + 5) = 2(6) = 12; multiple of 2, 3, 4, 6
P(2) = 2(2 + 1)(2 + 5) = 2(3)(7) = 42; multiple of 2, 3, 6, 7
P(3) = 3(3 + 1)(3 + 5) = 3(4)(8) = 96; multiple of 2, 3, 4, 6, 8, 12
..
Thus, from the above statements and verifying the options, we can say that n(n + 1)(n + 5) is a multiple of 3.
Q.12. The largest term in the expansion of (3+2x)^{50}, when x = ⅕ is
(a) 6^{th }term
(b) 7^{th} term
(c) 8^{th} term
(d) None of the above
Correct Answer is option (a)
The greatest term in the expansion of (x + y)^{n} is the kth term. Where k = [(n + 1)y]/[x + y]..(1)
On comparing the given expression with the general form, x = 3, y=2x, n=50
Now, substitute the values in the given expression, we get
Hence, kth term = [(50 + 1)(2x)]/[3 + 2x]
When x = ⅕,
K^{th} term = [(51)(2(⅕))]/[3 + 2(⅕)] = 6
Hence, the 6th term is the largest term in the expansion of (3 + 2x)^{50}, when x = ⅕.
Q.13. n^{2} < 2^{n} for all natural numbers
(a) n ≥ 5
(b) n < 5
(c) n > 1
(d) n ≤ 3
Correct Answer is option (a)
Consider, P(n) : n^{2} < 2^{n}
Substituting n = 1, 2, 3,…
P(1): 1^{2} < 2^{1}
1 < 2 (not true)
P(2): 2^{2} < 2^{2}
4 < 4 (not true)
P(3): 3^{2} < 2^{3}
9 < 8 (not true)
P(4): 4^{2} < 2^{4}
16 < 16 (not true)
P(5): 5^{2} < 2^{5}
25 < 32 (true)
P(6): 6^{2} < 2^{6}
26 < 64 (true)
Thus, n^{2} < 2^{n} for all natural numbers n ≥ 5.
Q.14. The coefficient of y in the expansion of (y^{2}+(c/y))^{5} is:
(a) 10c
(b) 29c
(c) 10c^{3}
(d) 20c^{3}
Correct Answer is option (c)
Given: (y^{2}+(c/y))^{5}
We know that general term of (x+a)^{n} is T_{r+1} =^{n}C_{r }x^{nr }a^{r}
Here, n=5, r=?
(y^{2}+(c/y))^{5} = ^{5}C_{r}.(y^{2})^{r}.(c/y)^{5r}
(y^{2}+(c/y))^{5} = 5Cr. y^{2r}. (c^{5r}/y^{5r})
On solving this, we get r = 2.
Hence, the coefficient of y = ^{5}C_{3}.c^{3} = 10c^{3}.
Therefore, option (c) 10c^{3} is the correct answer.
Q.15. If 10^{n} + 3.4^{n+2} + k is divisible by 9 for all n ∈ N, then the least positive integral value of k is
(a) 5
(b) 3
(c) 7
(d) 1
Correct Answer is option (a)
Given that 10^{n} + 3.4^{n+2} + k is exactly divisible by 9.
Consider: P(n) = 10^{n} + 3.4^{n+2} + k
Substituting n = 1,
P(1) = 10^{1 }+ 3.4^{1+2 }+ k
= 10 + 3(64) + k
= 10 + 192 + k
= 202 + k is exactly divisible by 9, the value of k will be 5.
Q.16. The fourth term in the expansion of (x2y)^{12} is:
(a) 1760 x^{9} × y^{3}
(b) 1670 x^{9} × y^{3}
(c) 7160 x^{9} × y^{3}
(d) 1607 x^{9} × y^{3}
Correct Answer is option (a)
We know that the general term of an expansion (a+b)^{n} is T_{r+1} = ^{n}C_{r} a^{nr }b^{r}.
Now, we have to find the fourth term in the expansion (x2y)^{12}
Hence, r = 3, a = x, b = 2y, n= 12.
Now, substitute the values in the formula, we get
T_{3+1} = ^{12}C_{3} x^{123 }(2y)^{3}.
On solving this, we get
T_{4} = 1760x^{9}y^{3}.
Q.17. Let P(n) : “2^{n} < (1 × 2 × 3 × … × n)”. Then the smallest positive integer for which P(n) is true is
(a) 1
(b) 2
(c) 3
(d) 4
Correct Answer is option (d)
P(1) : 2^{1} < 1
2< 1 is false
P(2) : 2^{2} < 1 × 2
4 < 2 is false
P(3) : 2^{3} < 1 × 2 × 3
8 < 6 is false
P(4) : 2^{4} < 1 × 2 × 3 × 4
16 < 24 is true
Q.18. If the fourth term of the binomial expansion of (px+(1/x))^{n} is 5/2, then
(a) n=6, p=6
(b) n=8, p=6
(c) n=8, p= ½
(d) n=6, p=½
Correct Answer is option (d)
Given: (px+(1/x))^{n}
Hence, the fourth term, T_{3+1 }= ^{n}C_{3}(px)^{n3}(1/x)^{3}
Given that the fourth term of the binomial expansion of (px+(1/x))^{n} is 5/2, which is independent of x.
Hence, (5/2)= ^{n}C_{3}(px)^{n3}(1/x)^{3} …(1)
On solving this, we get n=6.
Now, substitute n=6 in (5/2)= ^{n}C_{3}(p)^{3}
20p^{3} = 5/2
p^{3}=⅛
p=½.
Therefore, n=6 and p=½.
Q.19. For every positive integer n, 7n – 3n is divisible by
(a) 3
(b) 4
(c) 7
(d) 5
Correct Answer is option (b)
Let P(n) = 7^{n} – 3^{n}
Substituting n = 1, 2, 3,…
P(1) = 7^{1} – 3^{1} = 7 – 3 = 4
P(2) = 7^{2} – 3^{2} = 49 – 9 = 40
P(3) = 7^{3} – 3^{3} = 343 – 27 = 316
Thus, for every positive integer n, 7^{n} – 3^{n} is divisible by 4.
Q.20. If n is the positive integer, then 2^{3n} – 7n 1 is divisible by
(a) 7
(b) 10
(c) 49
(d) 81
Correct Answer is option (c)
Given: 2^{3n} – 7n 1. It can also be written as 8^{n} – 7n – 1
Let 8^{n} – 7n – 1 =0
So, 8^{n} = 7n+1
8^{n} = (1+7)^{n}
By applying binomial theorem, we get
8n – 1 – 7n = 49 (or) 2^{3n} – 7n 1 = 49
Hence, 2^{3n} – 7n 1 is divisible by 49.
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